Answer to Question #213709 in Linear Algebra for amit

Question

Show that if S and T are linear

transformations on a finite dimensional

vector space, then rank (ST)<= rank (S).

Answer 1

Let , $T:V\rightarrow V$ and $S:V\rightarrow V$ be two linear transformation from a finite dimensional vector space $V$ to itself.

Therefore, $S\circ T :V \rightarrow V$ is a linear transformation. Also note,

$dim(ImT)=rank(T), \\dim(ImS)=rank(S)$

and $dim(Im(ST))=dim(Im(S\circ T))=rank (ST).$

$dim(Ker (S))=nullity (S)\\ dim(ker(T))=nullity(T)$

and $dim(ker(ST))=nullity (ST)$ .

Note: Generally,$\ S\circ T$ written as $ST$ .

Now ,$Ker(ST)=\{x\in V :(ST)(x)=0 \}$ .

Where  is a zero vector of $V.$

$=\{ x\in V:S(T(x))=0 \}$

Now , if $x\in ker(T)$ then

$T(x)=0 \ \implies S(T(x))=S(0)=0$.

Hence, $x\in Ker(ST).$

Therefore,$Ker(T)\sube Ker(ST).$

Thus,$dim(Ker(T))\leq dim(Ker(ST))$

I,e,$nullity (T)\leq nullity (ST)...............(*)$

Now , from the rank-nullity theorem ,

$rank(T)+nullity(T)=rank(ST)+nullity (ST)=n$

Now , from (*) we conclude that ,

$rank(ST)\leq rank(T)$ .